Calculus is one of the substantial branches of mathematics deal with various important topics such as limits, integration, differentiation. |

If M(x) $\leq$ G(x) $\leq$ N(x) for all x in an open interval containing a, except possibly at a itself, and if

Lim$_{x \rightarrow a}$ M(x) = L = Lim$_{x \rightarrow a}$ N(x)

Then Lim$_{x \rightarrow a}$ G(x) exists and equal to L i.e.

Lim$_{x \rightarrow a}$ G (x) = L.

**To prove sequeeze theorem, consider the limit superior and inferior:**

L = Lim$_{x \rightarrow a}$M(x) $\leq$ Lim$_{x \rightarrow a}$ Inf$\leq$ G(x) $\leq$ Lim$_{x \rightarrow a}$SupG(x) $\leq$ Lim$_{x \rightarrow a}$N(x) = L

Consider ($\epsilon$, $\delta$), for all $\epsilon$ > 0 there exists $\delta$ > 0 such that

0 < |x - a| < $\delta$ (for all x)

=> -$\epsilon$ < G(x) - L < $\epsilon$

As Lim$_{x \rightarrow a}$ M(x) = L,

for all $\epsilon$, $\epsilon$ > 0 there exists $\delta_1$ > 0

and for all x, 0 < |x - a| < $\delta_1$

=> -$\epsilon$ < M(x) - L < $\epsilon$ ...(1)

Similarly, Lim$_{x \rightarrow a}$ N(x) = L,

for all $\epsilon$, $\epsilon$ > 0 there exists $\delta_2$ > 0

and for all x, 0 < |x - a| < $\delta_2$

=> -$\epsilon$ < N(x) - L < $\epsilon$ ...(2)

M(x) $\leq$ G(x) $\leq$ N(x)

or M(x) - L $\leq$ G(x) - L $\leq$ N(x) - L

If $\delta$ = min($\delta_1$, $\delta_2$), then if |x - a| < $\delta$

From equation (1) and (2), we obtain

-$\epsilon$ < M(x) - L $\leq$ G(x) - L $\leq$ N(x) - L < $\epsilon$

or -$\epsilon$ < G(x) - L < $\epsilon$

which shows that Lim$_{x \rightarrow a}$ G(x) = L.

**Few examples on the Squeeze theorem are given below:**

**Example 1:**Prove that Lim$_{y \rightarrow 0}$ y sin ($\frac{1}{y}$) = 0 using squeeze theorem.

**Solution:**Let G(y) = y sin ($\frac{1}{y}$) for y $\neq$ 0, then we have the inequalities

-|y| $\leq$ G(y) $\leq$ |y| ;for all y $\neq$ 0

Followed by well known inequality |Sin y| $\leq$ 1 $\forall$ y $\in$ R

=> Lim$_{y \rightarrow 0}$ (-|y| ) = 0 and Lim$_{y \rightarrow 0}$ (|y| ) = 0

By Squeeze theorem, Lim$_{y \rightarrow 0}$ y sin ($\frac{1}{y}$) = 0

**Example 2:**Let g(y) be a function s.t. |g(y)| $\leq$ L for any y $\neq$ 0. Prove that $lim_{y \rightarrow 0}$ yg(y) = 0

**Solution:**Since |g(y)| $\leq$ L for any y $\neq$ 0.

or |y g(y)| $\leq$ L|y|

=> -L|y| $\leq$ y g(y)| $\leq$ L|y|

For any non zero y,

$lim_{y \rightarrow 0}$ (-L|y|) = 0 = $lim_{y \rightarrow 0}$ (L|y|)

So by squeeze theorem, $lim_{y \rightarrow 0}$ yg(y) = 0

Squeeze theorem used to find the convergance and divergence of the sequences. If $lim_{n \rightarrow \infty}$ $a_n$ = $lim_{n \rightarrow \infty}$ $b_n$ = L and there exists an integer N s.t. $a_n$ $\leq$ $c_n$ $\leq$ $b_n$ for all n > N, then $\lim_{n \rightarrow \infty} $ = L.

**Let's understand this concept in a better with the help of an example:**

**Example:**Determine if the below sequence if converges or diverges.

{$b_n$ } = {$\frac{sin n}{n}$}

**Solution:**We know that, for all n, 1 $\leq$ sin n $\leq$ 1

Therefore, -$\frac{1}{n}$ $\leq$ $\frac{sin n}{n}$ $\leq$ $\frac{1}{n}$

Choose $a_n$ = -$\frac{1}{n}$ and $c_n$ = $\frac{1}{n}$

Also $a_n$ $\leq$ $c_n$ $\leq$ $b_n$ for all n

=> $lim_{n \rightarrow \infty}$$a_n$ = 0 = $lim_{n \rightarrow \infty}$$b_n$

So by sequeeze theorem $lim_{n \rightarrow \infty}$$c_n$ = 0

This implies, the sequence {$b_n$ } is converges to zero.